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Polar Double Integrals · Tutorial 570 of 999

Integrals over Intersections of Circular Regions

Translate intersecting circular regions into polar bounds, locate where the active radial boundary changes, and evaluate integrals using the resulting pieces.

Intermediate 10 min read

What You'll Learn

  • Convert an off-center circle equation into a polar radial boundary.
  • Decide when an off-center disk includes the origin and when rays meet an inner radial boundary.
  • Find angular transition points by solving for where two circle boundaries intersect.
  • Split an intersection integral wherever the smaller radial upper bound changes.
  • Check candidate bounds against the geometry and verify evaluated integrals.

Why Intersections Need Careful Bounds

When a region is the intersection of two disks, a point belongs to it only if it lies inside both disks. In polar coordinates, this usually means that the radial distance is limited by whichever disk boundary is reached first along a ray. Which boundary is reached first can change with the angle.

The main task is therefore not just to convert each circle into polar coordinates. It is to locate the angles where their radial boundaries meet, determine which bound is active between those angles, and split the integral accordingly. A correct split describes the same region on every angular interval without including points outside either disk.

Polar Bounds for a Disk

Let one disk have center \((a,b)\) and radius \(R\). Its Cartesian equation is \((x-a)^2+(y-b)^2\le R^2\). Using \(x=r\cos\theta\) and \(y=r\sin\theta\), the inequality becomes a quadratic in \(r\).

$$ r^2-2r(a\cos\theta+b\sin\theta)+a^2+b^2-R^2\le 0. $$

Writing \(c=a\cos\theta+b\sin\theta\), the two boundary roots, when real, are \(c\pm\sqrt{R^2-a^2-b^2+c^2}\). Along a given ray, the disk occupies the radial interval between these roots, subject to \(r\ge 0\). If the disk contains the origin, its lower root is nonpositive on every ray, so the ray's radial interval starts at zero.

For a disk centered at \((d,0)\), where \(d>0\), this simplifies to the boundary equation \(r^2-2dr\cos\theta+d^2=R^2\). When \(R>d\), the disk contains the origin, and its outer radial boundary is defined for every angle.

$$ r_{\mathrm{off}}(\theta) = d\cos\theta+\sqrt{R^2-d^2\sin^2\theta}. $$

Now intersect this disk with the disk centered at the origin of radius \(S\). The centered disk permits \(0\le r\le S\) at every angle. Since the off-center disk also contains the origin, the intersection has the radial bound

$$ 0\le r\le \min\{S,r_{\mathrm{off}}(\theta)\}. $$
Key rule: For disks that both contain the origin, the radial upper bound of their intersection is the smaller of their two radial upper bounds. The angular intervals must be split wherever those bounds exchange which one is smaller.

Finding the Transition Angles

A transition occurs when a ray meets both circle boundaries at the same radial distance. For the centered disk, that distance is \(r=S\). Substituting \(r=S\) into the off-center circle equation gives the transition condition directly.

$$ S^2-2dS\cos\theta+d^2=R^2, \qquad \cos\theta=\frac{S^2+d^2-R^2}{2dS}. $$

When \(d,S>0\), transition angles can occur only if the right-hand side lies in \([-1,1]\). If it lies strictly between \(-1\) and \(1\), there are generally two angles in a full revolution with that cosine. The equation identifies where the bounds are equal; test an angle between the transition points to decide which disk supplies the smaller bound there.

Important qualification: The minimum-bound description applies directly when the off-center disk contains the origin. If it does not, a ray that meets the disk has an interval between two positive roots, not an interval starting at zero. The intersection with a centered disk must then respect that inner radial boundary as well as the outer one. If the off-center disk is tangent to the origin, its rays of intersection are restricted to the half-plane facing its center.

Worked Example: Area of Two Intersecting Disks

Worked Example: Area of Two Intersecting Disks

Find the area of the intersection of the disk \(x^2+y^2\le 4\) and the disk \((x-2)^2+y^2\le 4\). The off-center disk is tangent to the origin, so its polar rays have \(-\pi/2\le\theta\le\pi/2\). Its equation becomes \(r^2-4r\cos\theta\le0\), or \(0\le r\le4\cos\theta\) on those rays. Together, the disks give \(0\le r\le\min\{2,4\cos\theta\}\).

Set the two candidate upper bounds equal: \(2=4\cos\theta\), so \(\cos\theta=1/2\). Within the allowed angular range the transition angles are \(\theta=\pm\pi/3\). At \(\theta=0\), \(4\cos\theta=4>2\), so the centered disk supplies the smaller bound between the transitions. Outside them, \(4\cos\theta<2\). The area integral, including the polar factor \(r\), is therefore

$$ A = \int_{-\pi/3}^{\pi/3}\int_0^2 r\,dr\,d\theta + \int_{-\pi/2}^{-\pi/3}\int_0^{4\cos\theta}r\,dr\,d\theta + \int_{\pi/3}^{\pi/2}\int_0^{4\cos\theta}r\,dr\,d\theta. $$

First, the central angular interval contributes

$$ \int_{-\pi/3}^{\pi/3}\left[\frac{r^2}{2}\right]_0^2\,d\theta = \int_{-\pi/3}^{\pi/3}2\,d\theta = \frac{4\pi}{3}. $$

The two outer contributions are equal, since \(\cos^2\theta\) is even. Combining them gives

$$ 2\int_{\pi/3}^{\pi/2}\left[\frac{r^2}{2}\right]_0^{4\cos\theta}\,d\theta = 16\int_{\pi/3}^{\pi/2}\cos^2\theta\,d\theta. $$

Using the antiderivative \(\theta/2+\sin(2\theta)/4\), the outer contribution is

$$ 16\left[ \frac{\theta}{2}+\frac{\sin(2\theta)}{4} \right]_{\pi/3}^{\pi/2} = 16\left(\frac{\pi}{4}-\frac{\pi}{6}-\frac{\sqrt{3}}{8}\right) = \frac{4\pi}{3}-2\sqrt{3}. $$

Thus the intersection area is \(A=8\pi/3-2\sqrt{3}\). As a check on the interval choice, the centered disk is the tighter constraint at \(\theta=0\), while the off-center disk is the tighter constraint at \(\theta=\pi/2\) and at \(\theta=-\pi/2\), where its radial extent is zero.

Why the Minimum and the Split Are Correct

For an angle at which both disks contain the origin, each disk includes every point on that ray from \(r=0\) up to its own outer boundary. A point lies in their intersection precisely when its radius does not exceed either boundary. That is equivalent to \(r\le\min\{S,r_{\mathrm{off}}(\theta)\}\).

The identity of the smaller boundary can change only where the two boundaries are equal. For the circle pair above, equality at a positive radius \(S\) requires the transition equation already derived. Between consecutive transition angles, the difference of the two continuous radial boundary functions cannot change sign without becoming zero. Consequently, testing one angle in each interval determines which boundary is smaller throughout that interval. This justifies splitting the integral at the transition angles and using one radial upper bound on each resulting piece.

Worked Example: Integrating \(x\) Over the Same Intersection

Worked Example: Integrating \(x\) Over the Same Intersection

Use the region from the previous example to evaluate \(\iint_D x\,dA\). In polar coordinates, \(x=r\cos\theta\) and \(dA=r\,dr\,d\theta\), so the integrand with its area factor is \(r^2\cos\theta\). The same three angular pieces apply.

On the central interval, the integral is

$$ \int_{-\pi/3}^{\pi/3}\int_0^2 r^2\cos\theta\,dr\,d\theta = \frac{8}{3}\int_{-\pi/3}^{\pi/3}\cos\theta\,d\theta = \frac{8\sqrt{3}}{3}. $$

For the two outer intervals together, symmetry gives

$$ \frac{2}{3}\int_{\pi/3}^{\pi/2}(4\cos\theta)^3\cos\theta\,d\theta = \frac{128}{3}\int_{\pi/3}^{\pi/2}\cos^4\theta\,d\theta. $$

An antiderivative of \(\cos^4\theta\) is \(3\theta/8+\sin(2\theta)/4+\sin(4\theta)/32\). Thus

$$ \int_{\pi/3}^{\pi/2}\cos^4\theta\,d\theta = \frac{3\pi}{16} - \left(\frac{\pi}{8}+\frac{7\sqrt{3}}{64}\right) = \frac{\pi}{16}-\frac{7\sqrt{3}}{64}. $$

The outer contribution is consequently \(8\pi/3-14\sqrt{3}/3\). Adding the central contribution gives

$$ \iint_D x\,dA = \frac{8\sqrt{3}}{3}+\frac{8\pi}{3}-\frac{14\sqrt{3}}{3} = \frac{8\pi}{3}-2\sqrt{3}. $$

As a check on the integration, the central term follows from \(\int_0^2 r^2\,dr=8/3\) and \(\int_{-\pi/3}^{\pi/3}\cos\theta\,d\theta=\sqrt{3}\). The outer term follows from the displayed antiderivative and the two equal outer intervals.

Worked Example: A Radial Integrand on a Tangent-Disk Intersection

Worked Example: A Radial Integrand on a Tangent-Disk Intersection

Let \(D\) be the intersection of \(x^2+y^2\le1\) and \((x-1)^2+y^2\le1\). Evaluate \(\iint_D 1/\sqrt{x^2+y^2}\,dA\). The integrand is undefined at the origin, but its polar form is \(1/r\) away from the origin, and the Jacobian factor \(r\) cancels it. The resulting radial integral is finite at zero.

The off-center disk gives \(0\le r\le2\cos\theta\) for \(-\pi/2\le\theta\le\pi/2\), while the centered disk gives \(r\le1\). The transition equation \(1=2\cos\theta\) gives \(\theta=\pm\pi/3\). Hence the centered bound applies in the central interval, and the off-center bound applies on the two outer intervals. The integral becomes

$$ \int_{-\pi/3}^{\pi/3}\int_0^1 1\,dr\,d\theta + 2\int_{\pi/3}^{\pi/2}\int_0^{2\cos\theta}1\,dr\,d\theta. $$

The central part is \(2\pi/3\). The outer part is

$$ 2\int_{\pi/3}^{\pi/2}2\cos\theta\,d\theta = 4[\sin\theta]_{\pi/3}^{\pi/2} = 4\left(1-\frac{\sqrt{3}}{2}\right) = 4-2\sqrt{3}. $$

The final value is \(2\pi/3+4-2\sqrt{3}\). To check, the central radial integral is \(\int_0^1 1\,dr=1\), and each outer radial integral is \(\int_0^{2\cos\theta}1\,dr=2\cos\theta\), exactly the expressions used above.

Common Pitfalls

A frequent error is to use the off-center circle's boundary as the upper radial bound at every angle, even where the centered disk is reached first. In the first example, at \(\theta=0\) that would give \(r\le4\), although the centered disk restricts \(r\le2\). The intersection instead has \(r\le\min\{2,4\}=2\) there.

Another error is to use \(r=4\cos\theta\) for the tangent disk at angles where \(\cos\theta<0\). For instance, at \(\theta=\pi\) this expression gives \(r=-4\), which is not a valid radial distance. The disk has no positive radial extent on that ray: its angular range is \(-\pi/2\le\theta\le\pi/2\), and its boundary formula applies only there.

Finally, equality of radial bounds identifies possible switching angles, but it does not by itself say which disk supplies the upper bound on either side. Test an angle in each interval. That check prevents a correct transition equation from being paired with incorrect pieces of the integral.

Check Your Understanding

Use the radial-bound and transition ideas to answer these questions.

  1. For the disk centered at \((d,0)\) with radius \(R\), write its polar boundary equation and its radial roots.
  2. Why does the intersection of two disks that both contain the origin use the smaller radial upper bound?
  3. For a centered disk of radius \(S\) and an off-center disk of radius \(R>d\) centered at \((d,0)\), derive the equation for a transition angle.
  4. In the intersection of \(x^2+y^2\le4\) and \((x-2)^2+y^2\le4\), which disk supplies the radial upper bound at \(\theta=0\), and which supplies it at \(\theta=\pi/2\)?
  5. Why must a disk tangent to the origin and centered at \((d,0)\) be treated as having a restricted angular range?

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